It is known that a linear endomorphism $T:V\to V$ on an $n$ dimensional $\mathbb{C}$-vector space can be decomposed according to the invariant subspaces of $V$ under $T$. This result can be proved as follows.

Consider $V$ as a $\mathbb{C}[x]$-module where $x$ acts on $v\in V$ by $Tv$ and use the convention $x^0=1$. Since $V$ is finite dimensional, there exists a non-trivial linear combination $P(T)$ of $\{1,T,...,T^n\}$ such that $P(T)$ annihilates $V$. Thus $V$ as a $\mathbb{C}[x]$-module is a finitely generated torsion module. Then we can apply the structure theorem of finitely generated modules over PIDs.

Does the above result have any implication to the space End$(V)$?

To me, I have thought about the following: the space End$(V)$ $\cong V^*\otimes V$ is a $n^2$ dimensional vector space. To study a member of it, the above result says instead of studying a linear combination of $n^2$ basis vectors, I can reduce to it some smaller combination if I make End$(V)$ into an algebra, though this reduction may only make sense when I want to study its invariant subspaces. Now this reduction seems as if a "quotient space", where I consider a map up to change of basis. But I can't make this idea clear to me.

I also wonder if there is a general framework to understand this kind of reduction (if it makes sense), as well as other such examples? Is there a notion of dimension in this algebra End$(V)$? Thanks in advance!

  • $\begingroup$ I'm not sure what you're saying in the second to last paragraph. What exactly do you mean by @reducing it to a smaller linear combination"? $\endgroup$ Apr 3, 2016 at 11:57
  • $\begingroup$ The algebra is question is simple and the decomposition into indecomposables as either a left- or right module over itself is well-known. $\endgroup$ Apr 3, 2016 at 12:32
  • $\begingroup$ Sorry for being not clear here. In general I am just wondering how would this structure theorem for a linear endomorphism $T$ help to understand the space of them i.e. End$V$. Does this make End$V$ any special? What I have written later is just classifying these endomorphisms according to their invariant subspaces, modulo basis change (i.e. classifying the Jordan canonical forms), in which I think I used the fact that two elements in End$V$ can be multiplied. Yet I don't know it is meaningful or not. $\endgroup$
    – Syl.Qiu
    Apr 3, 2016 at 12:55

1 Answer 1


Its quite obscure what you are asking for. But in everything you say I don't really see anything special to $End(V)$. Take $A$ any finite-dimensional $K$-algebra, and take $V$ any finite $A$-module. Then for any $a\in A$, you can make $V$ into a torsion $K[X]$-module through the action of $a$, and you will have a similar decomposition into $a$-invariant subspaces.

Of course you can say that in some sense this follows from the case of $End(V)$ since a finite $A$-module is nothing but a finite-dimensional $K$-vector space $V$ with a $K$-algebra morphism $A\to End(V)$.

As for your question about dimension, it's also unclear what you are asking for, but I can think of something that may interest you. First, you can entirely retrieve $V$ from the ring $A:= End(V)$. Indeed, $A$ is a central simple algebra over $K$, so from the ring $A$ you retrieve $K$ as its center, and then $V$ as the unique simple $A$-module. For any central simple algebra $A$ over $K$, you have $\dim_K A = n^2$ for some $n$, which is called the degree of $A$, and in the case $A=End(V)$, $\deg A = \dim_K V$.

If you want to talk about the dimension of subspaces, then you have the following fact : right ideals of $A$ are in canonical bijection with subspaces of $V$ : to $W\subset V$ you can associate $I = \{u\in A\,|\, Im(u)\subset W\}$. Then $End_K(W) = Hom_A(I)$, so the dimension of $W$ can be retrieved as $\deg Hom_A(I)$.

In this sense, you can speak of subspaces and their dimension purely in term of the ring $A = End(V)$.

  • $\begingroup$ Thanks for the answer! The part on central simple algebra looks indeed interesting! $\endgroup$
    – Syl.Qiu
    Apr 4, 2016 at 3:23

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .